Suppose \(Z(f) \ne \emptyset \) .
By [1]},
there exists a real analytic manifold \(M\) ,
a closed proper subset \(M^{\prime } \subset M\)  and a proper analytic map \(\phi : M \rightarrow U\)  so that \(\phi (M) = Z(f)\)  and \(d\phi (x)\)  is injective for each \(x \in M \setminus M^{\prime }\) .
As \(\dim U =2\)  then \(\dim M \in \lbrace 0,1,2\rbrace \) .
If \(\dim M = 0\)  then \(M\)  is a countable set of points and \(Z(f)\)  is countable.
If \(\dim M=2\)  then, by Theorem REF ,
there exists an open set \(O \subset U\)  where \(O \subset Z(f)\) ;
hence by Proposition REF ,
\(Z(f)=U\) .
So \(\dim M =1\) .
Now \(M\)  is the union of countably many sets \(\lbrace M_i : i \in I\rbrace \)  where each set \(M_i\)  is analytically diffeomorphic to either \(\mathbb {R}\)  or \(\mathbb {T}\) .
Since \(\operatorname{rank}d\phi (x) = 0\)  if and only if \(d \phi (x)\cdot d \phi (x) = 0\) ,
the set \(M^{\prime }\)  is the zero set of a non-zero analytic function with 1-dimensional domain.
The result now follows from Proposition REF .
